Question

In the following exercises, simplify using absolute value signs as needed.$\sqrt{242m^{23}},\sqrt[4]{405m^{10}},\sqrt[5]{160n^8}$

Short answer

The expression on simplification is $11\left|m^{11}\right|\sqrt{2m},3m^2\sqrt[4]{5m^2},2n\sqrt[5]{5n^3}$respectively

Step-by-step solution

Given information

Given the expression$\sqrt{242m^{23}},\sqrt[4]{405m^{10}},\sqrt[5]{160n^8}$

Using the product property

Using the product property, we get

$$\begin{gathered} \sqrt{242m^{23}}=\sqrt{121m^{22}·2m} \\ \sqrt{242m^{23}}=\sqrt{121m^{22}}·\sqrt{2m} \\ \sqrt{242m^{23}}=\sqrt{{\left(11m^{11}\right)}^2}·\sqrt{2m} \\ \sqrt{242m^{23}}==11\left|m^{11}\right|\sqrt{2m} \end{gathered}$$

Using the product property

Using the product property, we get

$$\begin{gathered} \sqrt[4]{405m^{10}}=\sqrt[4]{81m^8·5m^2} \\ \sqrt[4]{405m^{10}}=\sqrt[4]{81m^8}·\sqrt[4]{5m^2} \\ \sqrt[4]{405m^{10}}=\sqrt[4]{{\left(3m^2\right)}^4}·\sqrt[4]{5m^2} \\ \sqrt[4]{405m^{10}}=\left|3m^2\right|\sqrt[4]{5m^2} \\ \sqrt[4]{405m^{10}}=3m^2\sqrt[4]{5m^2} \end{gathered}$$

Using the product property

Using the property, we get

$$\begin{gathered} \sqrt[5]{160n^8}=\sqrt[5]{32n^5·5n^3} \\ \sqrt[5]{160n^8}=\sqrt[5]{32n^5}·\sqrt[5]{5n^3} \\ \sqrt[5]{160n^8}=\sqrt[5]{(2n{)}^5}·\sqrt[5]{5n^3} \\ \sqrt[5]{160n^8}=2n\sqrt[5]{5n^3} \end{gathered}$$